Dates in php Sunday and Saturday? - php

Question

i have one date.
example : $date='2011-21-12';
data format :yyyy-dd-mm;
IF date is Saturday or Sunday.
if Saturday add 2 day to the given date.
if Sunday add 1 day to the given date.
?

Answer 1

In a single line of code:
if (date('N', $date) > 5) $nextweekday = date('Y-m-d', strtotime("next Monday", $date));
If the day of week has a value greater than 5 (Monday = 1, Sat is 6 and Sun is 7) set $nextweekday to the YYYY-MM-DD value of the following Monday.
Editing to add, because the date format may not be accepted, you would need to reformat the date first. Add the following lines above my code:
$pieces = explode('-', $date);
$date = $pieces[0].'-'.$pieces[2].'-'.$pieces[1];
This will put the date in Y-m-d order so that strtotime can recognize it.

Answer 2

You can use the date and strtotime functions for this, like so:
$date = strtotime('2011-12-21');
$is_saturday = date('l', $date) == 'Saturday';
$is_sunday = date('l', $date) == 'Sunday';
if($is_saturday) {
$date = strtotime('+2 days', $date);
} else if($is_sunday) {
$date = strtotime('+1 days', $date);
}
echo 'Date is now ' . date('Y-m-d H:i:s', $date);

Answer 3

What about this:
$date='2011-21-12';
if (date("D", strtotime($date)) == "Sat"){
$new_date = date("Y-m-d", strtotime("+ 2 days",$date);
}
else if (date("D", strtotime($date)) == "Sun"){
$new_date = date("Y-m-d", strtotime("+ 1 day",$date);
}

Answer 4

The DateTime object can be really helpful for anything like this.
In this case.
$date = DateTime::createFromFormat('Y-d-m', '2011-21-12');
if ($date->format('l') == 'Sunday')
$date->modify('+1 day');
elseif ($date->format('l') == 'Saturday')
$date->modify('+2 days');
If you want to get the date back in that format.
$date = $date->format('Y-d-m');

Answer 5

$date = '2011-21-12'
$stamp = strtotime($date);
$day = date("l", $stamp);
if ($day == "Saturday"){
$stamp = $stamp + (2*+86400);
}elseif($day == "Sunday"){
$stamp = $stamp + 86400;
}
echo date("Y-d-m", $stamp);
The only reason i can think why this wouldnt work is strtotime not recognising that data format...

Answer 6

For sundays date('w',$date) will return 0, so the simplest test code is:
if(!date('w',strtotime($date)) { ... } //sunday

Answer 7

I just subtracted the difference from 8 an added it to the days.
if(date('N', strtotime($date)) >= 6) {
$n = (8 - date("N",strtotime($date)));
$date = date("Y-m-d", strtotime("+".$n." days");
}